Question

The chloride of a metal contains $71\% $ chlorine by weight and the vapour density of it is $50$. The atomic weigh of the metal will be:

• A. $29$
• B. $58$
• C. $35.5$
• D. $71$

Answer 1

Answer: (A)

Mol. wt. of metal chloride $=50 \times 2=100$
Let metal chloride be metal chlorine then
Equivalent of metal = Equivalents of chlorine
$=\frac{ W _{ m }}{ E _{ m }}=\frac{ W _{ Cl }}{ E _{ Cl }}$ or $\frac{29}{ E }=\frac{71}{35.5}$
$E=\frac{29}{2}$
Now $a+35.5 n=100$
or $n \times E +35 \times 5 n =100$
$\therefore n=2$
Therefore, $a=2 \times E=2 \times\left(\frac{29}{2}\right)=29$